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Fofino [41]
2 years ago
13

For the reaction 2NH3(g) + O2(g) -> 2NO(g) + 3H2O(g). The reaction is run using 824 g NH3 and excess oxygen. How many moles o

f NO are formed? How many moles of H2O are formed?
Chemistry
1 answer:
Alex73 [517]2 years ago
4 0

Answer:

Explanation:

Step One

Find the molar Mass of NH3

N =   14

3H =  3

Mass = 17

Step Two

Find the number of moles of NH3

mols = given mass / Molar Mass

given mass = 824 grams

molar mass = 17

mols = 824/17

mols =  48.47

Step Three

Find the moles of NO

Here you have to look at the balanced equation

Every 2 mols of HN3 produces 2 mols of NO

So the number of mols of NH3 = 48.47

The number of mols of  NO      = 48.47

Step Four

Find the mols of H2O

2 mols of NH3 produces 3 moles of H20

48.47 mols H20 produces x moles of H20

3/x = 2/48.47     Cross Multiply

2x = 3*48.47

2x = 145.41         Divide by 2

x = 145.41/2

x = 72.71            mols of water.

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The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
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Answer:

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Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

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Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

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At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

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